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Wed, 10 Sep 2008
Factorials are not quite as square as I thought
Let s(n) be the smallest perfect square larger than n. Then to have n! = a^{2} - 1 we must have a^{2} = s(n!), and in particular we must have s(n!) - n! square. This actually occurs for n in { 4, 5, 6, 7, 8, 9, 10, 11 }, and since 11 was as far as I got on the lunch line yesterday, I had an exaggerated notion of how common it is. had I worked out another example, I would have realized that after n=11 things start going wrong. The value of s(12!) is 21887^{2}, but 21887^{2} - 12! = 39169, and 39169 is not a square. (In fact, the n=11 solution is quite remarkable; which I will discuss at the end of this note.) So while there are (of course) solutions to 12! = a^{2} - b^{2}, and indeed where b is small compared to a, as I said, the smallest such b takes a big jump between 11 and 12. For 4 ≤ n ≤ 11, the minimal b takes the values 1, 1, 3, 1, 9, 27, 15, 18. But for n = 12, the solution with the smallest b has b = 288. Calculations with Mathematica by Mitch Harris show that one has n! = s(n!) - b^{2} only for n in {1, 4, 5, 6, 7, 8, 9, 10, 11, 13, 14, 15, 16}, and then not for any other n under 1,000. The likelihood that I imagine of another solution for n! = a^{2} - 1, which was already not very high, has just dropped precipitously. My thanks to M. Harris, and also to Stephen Dranger, who also wrote in with the results of calculations. Having gotten this far, I then asked OEIS about the sequence 1, 1, 3, 1, 9, 27, 15, 18, and (of course) was delivered a summary of the current state of the art in n! = a^{2} - 1. Here's my summary of the summary. The question is known as "Brocard's problem", and was posed by Brocard in 1876. No solutions are known with n > 7, and it is known that if there is a solution, it must have n > 10^{9}. According to the Mathworld article on Brocard's problem, it is believed to be "virtually certain" that there are no other solutions. The calculations for n ≤ 10^{9} are described in this unpublished paper of Berndt and Galway, which I found linked from the Mathworld article. The authors also investigated solutions of n! = a^{2} - b^{2} for various fixed b between 2 and 50, and found no solutions with 12 ≤ n ≤ 10^{5} for any of them. The most interesting was the 11! = 6318^{2} - 18^{2} I mentioned already. [ The original version of this article contained some confusion about whether s(n) was the largest square less than n, or the largest number whose square was less than n. Thanks to Roie Marianer for pointing out the error. ]
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